Chemistry - Solid State Question with Solution | TestHub

The density of a unit cell is given by the formula:

$d=\frac{Z\times M}{N_A\times a^3}$

Rearranging for molar mass $M$:

$M=\frac{d\times N_A\times a^3}{Z}$

For an fcc (face-centered cubic) structure, the number of atoms per unit cell is $Z=4$.

Given:
$d=10{\text{ g cm}}^{-3}$
$a=400\text{ pm}=400\times 10^{-10}\text{ cm}=4\times 10^{-8}\text{ cm}$
$N_A=6.022\times 10^{23}{\text{ mol}}^{-1}$

Calculating the volume of the unit cell $a^3$:

$a^3=(4\times 10^{-8}{)}^3=64\times 10^{-24}{\text{ cm}}^3$

Substituting the values into the formula:

$M=\frac{10\times 6.022\times 10^{23}\times 64\times 10^{-24}}{4}$

$M=\frac{10\times 6.022\times 10^{-1}\times 64}{4}$

$M=6.022\times 16$

$M=96.352{\text{ g mol}}^{-1}$

Answer: $96.35{\text{g mol}}^{-1}$